[Wien] specific heat factor
Ruben Weht
ruweht at cnea.gov.ar
Mon Jun 28 17:13:26 CEST 2004
Dear Peter,
I think Jan is right and there is an error in the gamma formula.
I think Gamma is (1/3).pi^2.kb^2.Na
where kb is the Bolztmann constant and Na Avogadro's number.
As Gamma is in general in units of mJ/k^2.mol
and density of states in states/Ry
you can take one kb in each energy unit.
In this way kb = 1.3806503.10^-20 mJ/K
and kb = 6.3336303.10^-6 Ry/K
Na = 6.0221.10^23 1/mol
So Gamma = 0.17324706 mJ.Ry/K^2/mol
I hope it helps,
Best,
Ruben
>> There is the electronic specific heat C_v/T, which is for
>> non-interacting electrons propotional to the density of states
>> at Fermi level, printed out in the output of tetra. I find the
>> convesion factor from N(E_F) in 1/Ry to gamma in mJ/(K^2 mol)
>> strange:
>>
>> gamma=N(E_F)/2.d0*(8.617D-5)
**2*pi**2)/3.d0*2.6255D6/27.212d0*1000.d0
>>
>> does anybody know where the 2.6255D6 comes from? By the way
>> conversion factor I normally use is
>> gamma[mJ/K^2/mol]=0.173*N(E_F)[1/Ry] (quite different from wien
>> value).
>
> Well, Georg seems to have programmed a "secret" here.
>
> To my understanding the conversions require k_B, Avogadros number
> and E-conversions into J
>
> 2.6255D6: This factor arises from two (or actually 3) numbers:
> Avogadros number (6.022 10^23 /mol)
> the energy conversion from "atomic units" to Joule, which is
>
> 27.212 * 1.60219 10^-19 (first converts hartree into eV, than
> eV into J)
>
> PS: The (8.617D-5)**2 is k_B in (eV/K)^2 and this leads to this
> additionl factor of 1/27.212 ....
>
> And somewhere are hidden the factors of 2 from Ry to hartree....
>
> Hope it fits like this. Just check it on a simple case (Cu or
> so), where the gamma0 is well known.
>
> P.Blaha
> -------------------------------------------------------------------
-------
> Peter BLAHA, Inst.f. Materials Chemistry, TU Vienna, A-1060
> Vienna Phone: +43-1-58801-15671 FAX:
> +43-1-58801-15698
> Email: blaha at theochem.tuwien.ac.at WWW:
> http://info.tuwien.ac.at/theochem/
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